About the IODS course

In this course, I will learn how to use R for statistical analysis.

My GitHub repository: https://github.com/jsimola/IODS-project

Week 2 - Regression and model validation

Read the data

rm(list = ls()) # clear workspace first
learning2014 <- read.csv("~/Documents/GitHub/IODS-project/learning2014.csv") # my own data wrangling
# read.table("~/Documents/GitHub/IODS-project/data/learning2014.txt", sep = ",") # Kimmo's data

Explore the structure and dimensions of the dataset

str(learning2014)
## 'data.frame':    166 obs. of  7 variables:
##  $ gender  : Factor w/ 2 levels "F","M": 1 2 1 2 2 1 2 1 2 1 ...
##  $ Age     : int  53 55 49 53 49 38 50 37 37 42 ...
##  $ Attitude: int  37 31 25 35 37 38 35 29 38 21 ...
##  $ deep    : num  3.58 2.92 3.5 3.5 3.67 ...
##  $ stra    : num  3.38 2.75 3.62 3.12 3.62 ...
##  $ surf    : num  2.58 3.17 2.25 2.25 2.83 ...
##  $ Points  : int  25 12 24 10 22 21 21 31 24 26 ...
dim(learning2014)
## [1] 166   7

The data includes 7 variables and 166 observations. The variables are: 1. Gender (Female = 1 Male = 2) 2. Age 3. Attitude 4-6. Mean scores of the deep, strategic and surface learning 7. Exam points

library(ggplot2) # Access the gglot2 library

Show a graphical overview of the data

# show gender distributions as bar graph
p1 <- ggplot(learning2014, aes(gender))
p1 + geom_bar()

# display variable distributions as histogram
p2 <- ggplot(learning2014, aes(Age))
p2 + geom_histogram(binwidth = 5)

p3 <- ggplot(learning2014, aes(Attitude))
p3 + geom_histogram(binwidth = 2)

p4 <- ggplot(learning2014, aes(deep))
p4 + geom_histogram(binwidth = 0.5)

p5 <- ggplot(learning2014, aes(stra))
p5 + geom_histogram(binwidth = 0.5)

p6 <- ggplot(learning2014, aes(surf))
p6 + geom_histogram(binwidth = 0.5)

p7 <- ggplot(learning2014, aes(Points))
p7 + geom_histogram(binwidth = 2)

# show relationships between variables 
p8 <- ggplot(learning2014, aes(x = Attitude, y = Points, col=gender))
p8 + geom_point() + ggtitle("Relationship between exam points and deep learning") + geom_smooth(method = "lm")

p9 <- ggplot(learning2014, aes(x = deep, y = Points, col=gender))
p9 + geom_point() + ggtitle("Relationship between exam points and deep learning") + geom_smooth(method = "lm")

p10 <- ggplot(learning2014, aes(x = stra, y = Points, col=gender))
p10 + geom_point() + ggtitle("Relationship between exam points and strategic learning") + geom_smooth(method = "lm")

p11 <- ggplot(learning2014, aes(x = surf, y = Points, col=gender))
p11 + geom_point() + ggtitle("Relationship between exam points and surface learning") + geom_smooth(method = "lm")

p12 <- ggplot(learning2014, aes(x = Age, y = Points, col=gender))
p12 + geom_point() + ggtitle("Relationship between age and exam points") + geom_smooth(method = "lm")

p13 <- ggplot(learning2014, aes(x = Age, y = Attitude, col=gender))
p13 + geom_point() + ggtitle("Relationship between age and attitudes") + geom_smooth(method = "lm")

library(GGally)
pairs(learning2014[-1], col = learning2014$gender)

p <- ggpairs(learning2014, mapping = aes(col=gender, alpha = 0.3), lower = list(combo = wrap("facethist", bins = 20)))

# draw the plot
p

Summary of the variables

summary(learning2014)
##  gender       Age           Attitude          deep            stra      
##  F:110   Min.   :17.00   Min.   :14.00   Min.   :1.583   Min.   :1.250  
##  M: 56   1st Qu.:21.00   1st Qu.:26.00   1st Qu.:3.333   1st Qu.:2.625  
##          Median :22.00   Median :32.00   Median :3.667   Median :3.188  
##          Mean   :25.51   Mean   :31.43   Mean   :3.680   Mean   :3.121  
##          3rd Qu.:27.00   3rd Qu.:37.00   3rd Qu.:4.083   3rd Qu.:3.625  
##          Max.   :55.00   Max.   :50.00   Max.   :4.917   Max.   :5.000  
##       surf           Points     
##  Min.   :1.583   Min.   : 7.00  
##  1st Qu.:2.417   1st Qu.:19.00  
##  Median :2.833   Median :23.00  
##  Mean   :2.787   Mean   :22.72  
##  3rd Qu.:3.167   3rd Qu.:27.75  
##  Max.   :4.333   Max.   :33.00
# another way of summarising the variables
#library(dplyr)
#learning2014 %>%
#  group_by(gender) %>%
#  summarise(mean = mean(Attitude), n = n())

The sample consist of mainly female participants. The majority of participants are between 20 to 30 years old. The variables are mostly normally distributed. The attitudes predict the exam points, but the learning strategies do not explain the exam points. Age does not explain the exam points or attitudes.

Fitting of a regression model to study whether attitudes, strategic and surface learning strategies explain the exam points. The learning strategies (strategic and surface) did not explain the exam points significantly. The exam points were significantly (p = 4.12e-09) explained by attitudes

# fit a linear model
my_model1 <- lm(Points ~ Attitude + stra + surf, data = learning2014) # how to use three explanatory vars
summary(my_model1)
## 
## Call:
## lm(formula = Points ~ Attitude + stra + surf, data = learning2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -17.1550  -3.4346   0.5156   3.6401  10.8952 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 11.01711    3.68375   2.991  0.00322 ** 
## Attitude     0.33952    0.05741   5.913 1.93e-08 ***
## stra         0.85313    0.54159   1.575  0.11716    
## surf        -0.58607    0.80138  -0.731  0.46563    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared:  0.2074, Adjusted R-squared:  0.1927 
## F-statistic: 14.13 on 3 and 162 DF,  p-value: 3.156e-08
my_model2 <- lm(Points ~ Attitude, data = learning2014) 
summary(my_model2)
## 
## Call:
## lm(formula = Points ~ Attitude, data = learning2014)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -16.9763  -3.2119   0.4339   4.1534  10.6645 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 11.63715    1.83035   6.358 1.95e-09 ***
## Attitude     0.35255    0.05674   6.214 4.12e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared:  0.1906, Adjusted R-squared:  0.1856 
## F-statistic: 38.61 on 1 and 164 DF,  p-value: 4.119e-09

Produce the diagnostic plots: Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage

plot(my_model2)

Week 3 - Logistic regression

Read and explore the data

rm(list = ls())  # clear workspace
alc <- read.csv("~/Documents/GitHub/IODS-project/data/alc.csv") # my own data wrangling
variable.names(alc) 
##  [1] "school"     "sex"        "age"        "address"    "famsize"   
##  [6] "Pstatus"    "Medu"       "Fedu"       "Mjob"       "Fjob"      
## [11] "reason"     "nursery"    "internet"   "guardian"   "traveltime"
## [16] "studytime"  "failures"   "schoolsup"  "famsup"     "paid"      
## [21] "activities" "higher"     "romantic"   "famrel"     "freetime"  
## [26] "goout"      "Dalc"       "Walc"       "health"     "absences"  
## [31] "G1"         "G2"         "G3"         "alc_use"    "high_use"

Student Performance Data Set Information

This data describes student performance in two Portuguese schools. The data attributes include school, student grades, demographic, social and school related features) and it was collected by using school reports and questionnaires. The data were combined from two datasets: (1) performance in mathematics and (2) portuguese language. The grades are means of Math and Portuguese: G1 - first period grade (numeric: from 0 to 20) G2 - second period grade (numeric: from 0 to 20) G3 - final grade (numeric: from 0 to 20, output target)

Hypotheses of the relationship between high/low alcohol consumption and four other variables

  • H1: final grade (G3) is higher for low alcohol consumers
  • H2: high alchohol use is associated with elevated number of school absences
  • H3: low alcohol use is associated with better health
  • H4: low alchohol use is associated with longer study time per week

The distributions of the chosen variables and their relationships with alcohol consumption

# access the tidyverse libraries tidyr, dplyr, ggplot2
library(tidyr); library(dplyr); library(ggplot2)
## 
## Attaching package: 'dplyr'
## The following object is masked from 'package:GGally':
## 
##     nasa
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
chosen_vars <- c("high_use","G3","absences","health","studytime") # choose relevant vars
chosen_data <- select(alc, one_of(chosen_vars))
#chosen_data$high_useN <- as.numeric(chosen_data$high_use)
gather(chosen_data) %>% glimpse
## Observations: 1,910
## Variables: 2
## $ key   <chr> "high_use", "high_use", "high_use", "high_use", "high_us...
## $ value <int> 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,...
gather(chosen_data) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar() # bar plots

# box plots
ggplot(chosen_data, aes(x=high_use, y = G3)) + geom_boxplot()

ggplot(chosen_data, aes(x=high_use, y = absences)) + geom_boxplot()

ggplot(chosen_data, aes(x=high_use, y = health)) + geom_boxplot()

ggplot(chosen_data, aes(x=high_use, y = studytime)) + geom_boxplot()

Distibutions:

  • The distibution of school absences is right skewed with only a few students having more absences of more than 10, while most students are between 0-5.
  • Final grade is almost normally distributed.
  • Healt is left skewed, because most students have very good health (= 5).
  • Low alcohol consumption is more common than high alcohol use (approx 25 %).
  • Study time is almost normally distributed.

The effect of alcohol consumption on the selected variables:

  • Final grade is better for low alcohol consumers as suggested by H1.
  • The number of school absences is higher for those who use more alcohol as suggested by H2.
  • No difference in health between high and low alchohol consumers, contrary to H3.
  • Study time is longer for those who have low alcohol consumption as suggested by H4.

Using logistic regression to statistically explore the relationship between the chosen variables and alcohol consumption.

m <- glm(high_use ~ G3 + absences + health + studytime, data = chosen_data, family = "binomial")  # find the model with glm()
summary(m) # summary of the model
## 
## Call:
## glm(formula = high_use ~ G3 + absences + health + studytime, 
##     family = "binomial", data = chosen_data)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2434  -0.8399  -0.6601   1.1619   2.1430  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  0.11336    0.61876   0.183 0.854635    
## G3          -0.05307    0.03602  -1.473 0.140698    
## absences     0.07837    0.02274   3.446 0.000569 ***
## health       0.06701    0.08551   0.784 0.433240    
## studytime   -0.50608    0.15731  -3.217 0.001295 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 465.68  on 381  degrees of freedom
## Residual deviance: 430.62  on 377  degrees of freedom
## AIC: 440.62
## 
## Number of Fisher Scoring iterations: 4
OddsRatios <- coef(m) %>% exp
ConfInt <- confint(m) %>% exp
## Waiting for profiling to be done...
cbind(OddsRatios, ConfInt)
##             OddsRatios     2.5 %    97.5 %
## (Intercept)  1.1200365 0.3308837 3.7712228
## G3           0.9483123 0.8833403 1.0177770
## absences     1.0815248 1.0366149 1.1334569
## health       1.0693034 0.9055302 1.2671462
## studytime    0.6028539 0.4387386 0.8140308

Interpretation of the results:

  • The odds ratio (probability of high alcohol use) for absences and health are higher than for the final grade and study time.
  • Curiously, however, the coeffients for absences and study time are significant (at p < 0.001 and p < .01, respectively).
  • Absences and alcohol use show a positive relationship. This result is in line with H2 that predicted more absences for those who consume more alcochol.
  • The relationship between study time and alcohol use is negative and thus follows the prediction of H4. That is, low alcohol use is associated with longer study time.
  • Logistic regression showed that alcohol use is not related to the final grade (G3) and health.

Selecting absences and studytime that had a statistical relationship with alcohol consumption.

m2 <- glm(high_use ~ absences + studytime, data = chosen_data, family = "binomial") # find the model with glm()
summary(m2) 
## 
## Call:
## glm(formula = high_use ~ absences + studytime, family = "binomial", 
##     data = chosen_data)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2128  -0.8387  -0.7046   1.1996   2.1832  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -0.16640    0.33954  -0.490 0.624087    
## absences     0.08054    0.02285   3.524 0.000425 ***
## studytime   -0.55015    0.15550  -3.538 0.000403 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 465.68  on 381  degrees of freedom
## Residual deviance: 433.65  on 379  degrees of freedom
## AIC: 439.65
## 
## Number of Fisher Scoring iterations: 4
# predict() the probability of high_use
probabilities <- predict(m2, type = "response")

# add the predicted probabilities to 'chosen_data'
chosen_data <- mutate(chosen_data, probability = probabilities)
chosen_data <- mutate(chosen_data, prediction = probability > 0.5)

# 2x2 cross tabulation of predictions versus the actual values
t <- table(high_use = chosen_data$high_use, prediction = chosen_data$prediction)
t
##         prediction
## high_use FALSE TRUE
##    FALSE   256   12
##    TRUE     96   18
# The total proportion of inaccurately classified individuals (= the training error) 
NumErr = t[2,1] + t[1,2] 
Tot = 382 # num obs
PropErr = round(NumErr / Tot, 2)
Accuracy = 100 - (PropErr*100)

The model has high predictive power. In total, the model achieved an accurcay of 72% in correct classification of individuals to high/low alcohol consumers based on their school absences and study time. This much better than guessing which would give an accuracy of 50%.

Week 4 - Clustering and classification

rm(list = ls())  # clear workspace
library(MASS)
## 
## Attaching package: 'MASS'
## The following object is masked from 'package:dplyr':
## 
##     select
data("Boston")
str(Boston)
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

Boston data set

This data has 506 osbervations (rows) and 14 variables (columns). The data includes information about housing values of suburbs of Boston, such as crime rate, business, accessibility to highways, and characteristics of the citizens in the area.

Plot the distibutions of the variables

gather(Boston) %>% glimpse
## Observations: 7,084
## Variables: 2
## $ key   <chr> "crim", "crim", "crim", "crim", "crim", "crim", "crim", ...
## $ value <dbl> 0.00632, 0.02731, 0.02729, 0.03237, 0.06905, 0.02985, 0....
gather(Boston) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_histogram() # bar plots
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

  • ‘age’, ‘black’ and ‘ptratio’ are left skewed
  • ‘dis’, ‘Istat’ and ‘nox’ are right skewed
  • most values of ‘chas’, ‘crim’ and ‘zn’ are zeros
  • ‘indus’, ‘rad’ and ‘tax’ are bimodal
  • ‘medv’ and ‘rm’ roughly normally distributed

Summarize data

summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

Correlations between the variables

cor_matrix <-cor(Boston) 
cor_matrix <-round(cor_matrix, digits=2)
cor_matrix 
##          crim    zn indus  chas   nox    rm   age   dis   rad   tax
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47
##         ptratio black lstat  medv
## crim       0.29 -0.39  0.46 -0.39
## zn        -0.39  0.18 -0.41  0.36
## indus      0.38 -0.36  0.60 -0.48
## chas      -0.12  0.05 -0.05  0.18
## nox        0.19 -0.38  0.59 -0.43
## rm        -0.36  0.13 -0.61  0.70
## age        0.26 -0.27  0.60 -0.38
## dis       -0.23  0.29 -0.50  0.25
## rad        0.46 -0.44  0.49 -0.38
## tax        0.46 -0.44  0.54 -0.47
## ptratio    1.00 -0.18  0.37 -0.51
## black     -0.18  1.00 -0.37  0.33
## lstat      0.37 -0.37  1.00 -0.74
## medv      -0.51  0.33 -0.74  1.00
corrplot(cor_matrix, method = "color")

Overall, the variables show strong correlations with each other, therefore only the strongest (> 0.7 or < -0.7) are interpreted below

  • ‘indus’ correlates positively with ‘nox’ (0.76) and ‘tax’ (0.72) and negatively with ‘dis’ (-0.71), suggesting that the proportion of non-retail business acres is associated with higher nitrogen oxides concentration and tax rate and with shorter distances from Boston employment centres.

  • ‘nox’ correlates positively with ‘age’ (0.73) and negatively with ‘dis’ (-0.77), suggesting that the proportion of units built prior to 1940 is associated with higher nitrogen oxides concentration and shorter distance from employment centres.

  • ‘rm’ correlates positively with ‘medv’ (0.7), suggesting that the average number of rooms is associated with higher median of owner-occupied homes in $1000s.

  • ‘age’ correlates negatively with ‘dis’ (-0.75), suggesting that the proportion of units built prior to 1940 is associated with shorter distance from employment centres.

  • ‘rad’ correlates with ‘tax’ (0.91): accessibility to highways is highly correlated with tax rate.

  • ‘lstat’ correlates negatively with ‘medv’ (-0.74): the percentage of lower status population (is this appropriate??) is associated with lower amount of owner-occupied homes in $1000s.

Standardize the data

boston_scaled <- scale(Boston)
summary(boston_scaled)
##       crim                 zn               indus        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202  
##       chas              nox                rm               age         
##  Min.   :-0.2723   Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331  
##  1st Qu.:-0.2723   1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366  
##  Median :-0.2723   Median :-0.1441   Median :-0.1084   Median : 0.3171  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.:-0.2723   3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059  
##  Max.   : 3.6648   Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164  
##       dis               rad               tax             ptratio       
##  Min.   :-1.2658   Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047  
##  1st Qu.:-0.8049   1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876  
##  Median :-0.2790   Median :-0.5225   Median :-0.4642   Median : 0.2746  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6617   3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058  
##  Max.   : 3.9566   Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372  
##      black             lstat              medv        
##  Min.   :-3.9033   Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.: 0.2049   1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median : 0.3808   Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.4332   3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 0.4406   Max.   : 3.5453   Max.   : 2.9865

The data is now standardized to zero mean.

Create a categorical variable of the crime rate

boston_scaled <- as.data.frame(boston_scaled)

bins <- quantile(boston_scaled$crim) # create a quantile vector of crim and print it

# create a categorical variable 'crime' 
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))
table(crime) # print the table
## crime
##      low  med_low med_high     high 
##      127      126      126      127
boston_scaled <- cbind(boston_scaled, crime) # bind crime to boston_scaled
boston_scaled <- dplyr::select(boston_scaled, -crim) # remove original crim from the dataset

Divide the dataset to train (80%) and test (20%) sets

n <- nrow(boston_scaled) # number of rows
ind <- sample(n,  size = n * 0.8) # choose randomly 80% of the rows
train <- boston_scaled[ind,] # create the train set
test <- boston_scaled[-ind,] # create the test set

Fitting a linear discriminant analysis (LDA) to the dataset

lda.fit <- lda(crime  ~., data = train) # crime rate is the target variable and all the other variables are predictors 
lda.fit # print the solution
## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2623762 0.2524752 0.2475248 0.2376238 
## 
## Group means:
##                   zn      indus         chas        nox         rm
## low       0.95945867 -0.8736810 -0.123759247 -0.8608359  0.4715759
## med_low  -0.08137866 -0.2445444 -0.002135914 -0.5760720 -0.1233722
## med_high -0.38261997  0.1435672  0.239493963  0.3054509  0.2108483
## high     -0.48724019  1.0172418 -0.108283225  1.0174409 -0.3976487
##                 age        dis        rad        tax     ptratio
## low      -0.8393850  0.8103087 -0.6839198 -0.7576931 -0.44223764
## med_low  -0.3631045  0.3797533 -0.5495071 -0.4572909 -0.04240666
## med_high  0.3826833 -0.3195191 -0.3800745 -0.3092325 -0.28847543
## high      0.7706830 -0.8292402  1.6368728  1.5131579  0.77931510
##               black       lstat         medv
## low       0.3809885 -0.76048984  0.561007732
## med_low   0.3114052 -0.13188358 -0.004846203
## med_high  0.1088360 -0.07204296  0.263799741
## high     -0.8592553  0.86339861 -0.722996383
## 
## Coefficients of linear discriminants:
##                 LD1         LD2         LD3
## zn       0.08238760  0.84587350 -0.83645794
## indus    0.00447987 -0.17532951  0.28064174
## chas    -0.08432467 -0.10650700  0.07333207
## nox      0.40629109 -0.71536542 -1.30313753
## rm      -0.11596422 -0.06941634 -0.14668263
## age      0.25580500 -0.32138650 -0.25265126
## dis     -0.04862418 -0.46286056  0.17841191
## rad      2.99045701  0.95066999 -0.26396363
## tax     -0.01498848 -0.21052792  0.73536274
## ptratio  0.08595505  0.05934535 -0.21134140
## black   -0.15467486  0.00223080  0.08973001
## lstat    0.23255642 -0.23753552  0.42215818
## medv     0.19459575 -0.44943279 -0.21178783
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9457 0.0384 0.0159
classes <- as.numeric(train$crime) # target classes as numeric
plot(lda.fit, dimen = 2, col = classes, pch = classes) # plot the lda results

Predict the classes with the LDA model on the test data

correct_classes <- test$crime # save the crime categories from the test set 
test <- dplyr::select(test, -crime) # remove crime from the test dataset
lda.pred <- predict(lda.fit, newdata = test) # predict classes with test data
table(correct = correct_classes, predicted = lda.pred$class) # Cross tabulate the results 
##           predicted
## correct    low med_low med_high high
##   low       15       6        0    0
##   med_low    4      18        2    0
##   med_high   0       6       20    0
##   high       0       0        0   31

Overall, the model performs well in predicting the crime categories. The high crime category is predicted fully, medium high is predicted well execpt small confusion with medium low crime rate category. Low and med_low categories are mixed.
### Reload and normalize Boston dataset and calculate the euclidean distances between the observations

library(MASS)
data('Boston')
boston_scaled <- scale(Boston)
boston_scaled <- as.data.frame(boston_scaled)
dist_eu <- dist(boston_scaled) #Calculate the  distances

Cluster the dataset with k-means function and plot the clusters. First, investigate what is the optimal number of clusters by looking at how the sum of squared distance changes with the number of clusters

k_max <- 25 # determine the maximum number of clusters
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss}) # calculate the total within sum of squares
qplot(x = 1:k_max, y = twcss, geom = 'line') # visualize the results

km <-kmeans(boston_scaled, centers = 5) # k-means clustering
pairs(boston_scaled, col = km$cluster) # plot the clusters

Five seems to be an optimal number of clusters, because a drop in the sum of squared distances around there.

library(MASS)
data('Boston')
boston_scaled <- scale(Boston)
boston_scaled <- as.data.frame(boston_scaled)
# Perform k-means on the original Boston data with some reasonable number of clusters (> 2)
km_4 <-kmeans(boston_scaled, centers = 4) # k-means clustering
clusters <- as.numeric(km_4$cluster)
lda.fit4 <- lda(clusters  ~., data = boston_scaled) # perform LDA using the clusters  
lda.fit4
## Call:
## lda(clusters ~ ., data = boston_scaled)
## 
## Prior probabilities of groups:
##         1         2         3         4 
## 0.3478261 0.1996047 0.1916996 0.2608696 
## 
## Group means:
##         crim         zn      indus        chas        nox          rm
## 1 -0.3955833 -0.2254704 -0.5890095 -0.11573985 -0.6008751 -0.06953428
## 2 -0.3961410  1.4923250 -1.0073590  0.03952046 -0.8201975  1.14916783
## 3 -0.3166884 -0.4817148  0.7363879  0.21473488  0.5617886 -0.56201562
## 4  1.0632703 -0.4872402  1.0149946 -0.03371693  1.0159127 -0.37357880
##          age        dis        rad        tax     ptratio       black
## 1 -0.4918015  0.4322643 -0.5792552 -0.6830752 -0.01226651  0.36362199
## 2 -0.9067300  0.9150799 -0.5986698 -0.6556098 -1.01943385  0.35076234
## 3  0.8101020 -0.6166607 -0.5840517 -0.1592236 -0.01279411 -0.00268091
## 4  0.7542188 -0.8233749  1.6596029  1.5294129  0.80577843 -0.75124560
##        lstat        medv
## 1 -0.4012240  0.07402657
## 2 -0.9782213  1.20570299
## 3  0.6131707 -0.48287523
## 4  0.8328654 -0.66640742
## 
## Coefficients of linear discriminants:
##                 LD1         LD2         LD3
## crim     0.08139384 -0.22359028  0.06554147
## zn       0.33554601 -0.88266028 -1.21807881
## indus    0.33981045  0.59239693 -0.66723420
## chas    -0.01718605  0.12636697 -0.09003424
## nox     -0.01896734 -0.03907779 -0.36248494
## rm      -0.06514429 -0.54171225 -0.27478691
## age      0.04189907  0.47483313 -0.49074354
## dis     -0.22488672 -0.10703816  0.20753920
## rad      5.99985509 -1.25972869  1.35203394
## tax      0.22488689  0.01764947 -0.74090656
## ptratio  0.25626572  0.24911107 -0.03140773
## black   -0.04205519 -0.04273764  0.09403661
## lstat    0.02170577 -0.15538625 -0.50478462
## medv    -0.11487427 -0.61561036 -0.39927321
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9131 0.0678 0.0191
# the function to draw lda biplot arrows 
my_scale = 10 # unable to draw the arrows - i'm not sure what myscale does -tested with different values without any effect 
lda.arrows <- function(x, myscale = my_scale, arrow_heads = 0.1, color = "black", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

plot(lda.fit4, dimen = 3, col = clusters, pch = clusters)
lda.arrows(lda.fit4, myscale = my_scale)